reinforced concrete design

REINFORCED CONCRETE DESIGN

Dr. AHMED AJEL

Strength Design Method 3.1 Strength Design Method The Strength Design Method requires that the design strength of a member at any section should equal or exceed the required strength calculated by the code-specified factored load combinations. In general,

Design Strength ≥ Required Strength (U) Where Design Strength = Strength Reduction Factor (ϕ) x Nominal Strength ϕ = Strength reduction factor that accounts for 1- the probability of understrength of a member due to variations in material strengths and dimensions. 2- inaccuracies in the design equations. 3- the degree of ductility and required reliability of the loaded member. 4-the importance of the member in the structure. Nominal Strength = Strength of a member or cross-section calculated using assumptions and strength equations of the Strength Design Method before application of any strength reduction factors. Required Strength (U) = Load factors x Service load effects. The required strength is computed in accordance with the load combinations. Load Factor = Overload factor due to probable variation of service loads. Service Load = Load specified by general building code (unfactored).

Notations Required strength: Mu = factored moment (required flexural strength) Pu = factored axial force (required axial load strength) Vu = factored shear force (required shear strength) Tu = factored torsional moment (required torsional strength) Nominal strength: Mn = nominal moment (required flexural strength) Pn = nominal axial force (required axial load strength) Vn = nominal shear force (required shear strength) Tn = nominal torsional moment (required torsional strength) 1

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Design Strength: Φ Mn = design moment (required flexural strength) Φ Pn = design axial force (required axial load strength) Φ Vn = design shear force (required shear strength) Φ Tn = design torsional moment (required torsional strength) 3.2 Loads Loads that act on structures can be divided into three broad categories : dead loads , live loads , and environmental loads.

3.2.1 Dead Load Dead loads consist of the weight of all materials of construction incorporated into the building including, but not limited to, wall, floors, roofs, ceilings, stairways, built-in partition, finishes , cladding, and other similarly incorporated architectural and structural items, and fixed service equipment including the weight of cranes. Dead loads can be determined with high degree of precision, although not until after the structure has been designed. For this reason it is necessary to estimate dead load before a structure analysis is made, in order that the part of the internal forces due to weight can be taken into account. 2

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Density of Building Materials Building Materials kN/m2 Brick 18.63 Gypsum 11.77 Light Weight Concrete 16.67 Plain Concrete 22.65 Reinforced Concrete 23.54 Dry Soil 15.7 Hollow Block 13.73 Sand 15.7 Steel 76.99 Thermoston 8.83

3.2.2 Live Load Live loads consist chiefly or partially of occupancy loads in buildings and traffic loads on bridges. They may be either fully or partially in place or not present at all, and may also change in location. Live load may be steady or unsteady; they may be fixed, movable, or moving; they may be applied slowly or suddenly. The live loads which usually be considered are: 1- The weight of people, furniture, and machinery. 2- The weight of traffic on a bridge. 3- The weight of snow. 4- Dynamic forces resulting from moving loads.

3.2.3 Environmental Loads Environmental loads consist mainly of snow loads, wind pressure and suction, earthquake loads, soil pressure on subsurface portions of structures, loads from possible ponding of rainwater on flat surfaces, and forces caused by temperature differentials. Like live loads , environmental loads at any given time are uncertain in both magnitude and distributions.

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Minimum Distributed Live Load 1

2

3

Occupancy or Use Libraries Reading rooms Stack Rooms Corridors above firstfloor School Class rooms

kN/m2

First-floor corridors

4.79

6 2.87 7.18 3.83 7 1.92

Corridors above first3.83 floor Stadiums and arenas

8

Occupancy or Use Hospitals Operation room, laboratories Patient rooms Corridors above first floor

kN/m2 2.87 1.92 3.83

Hotels and multifamily houses Private rooms and corridors 1.92 serving them Public rooms and corridors 4.79 serving them Office Buildings

Lobbies and first-floor 4.79 corridors Bleachers 4.79 Offices 2.4 Fixed seats 2.87 Corridors above first-floor 3.83 4 9 Storage warehouses Residential, (one and two family) Light 6.0 Uninhabitable attics without 0.48 storage Heavy 11.97 Uninhabitable attics with 0.96 storage 5 Habitable attics and sleeping 1.92 Stores areas First floor 4.79 10 Manufacturing Upper floors 3.59 Light 6.00 Wholesale, all floors 6.00 Heavy 11.97 ASCE Standard ,2005, "Minimum Design Loads for Buildings and Other Structure"

3.3 Required Strength (Load Factors) As previously stated, the required strength U is expressed in terms of factored loads, or their related internal moments and forces. Factored loads are the service-level loads specified in the general building code, multiplied by appropriate load factors in ACI Code section 9.2. Load factors are numbers, almost larger than 1.0, which are used to increase the estimated loads applied to structures. They are used for loads applied to all types of members, not just beams and slabs. The loads are increased to attempt to account for uncertainties involved in estimating their magnitude The ACI Code Section 9.2 prescribes load factors for specific combinations of loads. A list of these combinations is shown below. The numerical value of the load factor assigned to each type of load is influenced by the degree of accuracy with which the load can usually be 4

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assessed, the variation which may be expected in the load during the lifetime of a structure and the probability of simultaneous occurrence of different load types. Hence, dead loads, because they can usually be more accurately determined and are less variable, are assigned a lower load factor (1.2) as compared to live loads (1.6). Also, weight and pressure of liquids with well-defined densities and controllable maximum heights are assigned a reduced load factor of 1.2 due the lesser probability of overloading. A higher load factor of 1.6 is required for earth and groundwater pressures due to considerable uncertainty of their magnitude and recurrence. Note that while most usual combinations of loads are included, it should not be assumed that all cases are covered. Section 9.2 contains load combination as follows:

U = 1.4(D + F)…………………………………………………..…. (9-1) U = 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)…………….. (9-2) U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W)…………………... (9-3) U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)……………………… (9-4) U = 1.2D + 1.0E + 1.0L + 0.2S ………….…………………….……(9-5) U = 0.9D + 1.6W + 1.6H …………………………………….……..(9-6) U = 0.9D + 1.0E + 1.6H ……………………………………………(9-7)

Where: D = dead loads, or related internal moments and forces. E = load effects of seismic forces, or related internal moments and forces. F = loads due to weight and pressures of fluids with well-defined densities and controllable maximum heights, or related internal moments and forces. H = loads due to weight and pressure of soil, water in soil, or other materials, or related internal moments and forces. L = live loads, or related internal moments and forces. Lr = roof live load, or related internal moments and forces. R = rain load, or related internal moments and forces. S = snow load, or related internal moments and forces. T =cumulative effect of temperature, creep, shrinkage, differential settlement, and shrinkage compensating concrete. U = required strength to resist factored loads or related internal moments and forces. W = wind load, or related internal moments and forces.

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Example 3.1 The axial loads for a building column have been estimated with the following results: D = 150 kN, live load from roof Lr = 60 kN, live loads from floors L = 300 kN, compression wind W = 70 kN, tensile wind W= 60 kN, seismic compression load = 50 kN, and tensile seismic load = 40 kN. Determine critical design load using the ACI load combinations. Solution (9-1) U = (1.4)(150 + 0) = 210 kN (9-2) U = (1.2)(150 +0+0) +(1.6)(300 + 0) + (0.5)(60) = 690 kN (9-3)(a) U = (1.2)(150) + (1.6)(60) + (1.0)(300) =576 kN (b) U = (1.2)(150) + (1.6)(60) + (0.8)(70) = 332 kN (c) U = (1.2)(150) + (1.6)(60) + (0.8)(-60) = 228kN (9-4)(a) U = (1.2)(150) + (1.6)(70) + (1.0)(300) + (0.5)(60) = 622 kN (b) U = (1.2)(150) + (1.6)(-60) + (1.0)(300) + (0.5)(60) = 414 kN (9-5)(a) U = (1.2)(150) + (1.0)(50) + (1.0)(300) + (0.2)(0) = 530 kN (b) U = (1.2)(150) + (1.0)(-40) + (1.0)(300) + (0.2)(0) = 440 kN (9-6)(a) U = (0.9)(150) + (1.6)(70) +(1.6)(0) = 247 kN (b) U = (0.9)(150) + (1.6)(-60) + (1.6) (0) = 39 kN (9-7)(a) U = (0.9)(150) + (1.0)(50) + (1.6) (0) = 185 kN (b) U = (0.9)(150) + (1.0)(-40) + (1.6) (0) = 95 kN Answer : Largest value = 690 kN.

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3.4 ACI Code Provisions (Design Strength) The design strength provided by a member, its connections to other members, and its cross-section, in terms of flexure, axial load, shear, and torsion, is equal to the nominal strength calculated in accordance with the provisions and assumptions stipulated in the code, multiplied by a strength reduction factor ϕ, which is less than unity. The rules for computing the nominal strength are based generally on conservatively chosen limit states of stress, strain, cracking or crushing, and conform to research data for each type of structural action. The ACI Code provided for these variables by using the following ϕ factors as provided below : 9.3.1 - Design strength provided by a member, its connections to other members, and its cross sections, in terms of flexure, axial load, shear, and torsion, shall be taken as the nominal strength calculated in accordance with requirements and assumptions of this Code, multiplied by the strength reduction factors ϕ in 9.3.2, 9.3.4, and 9.3.5. 9.3.2 - Strength reduction factor ϕ shall be as given in 9.3.2.1 through 9.3.2.7: 9.3.2.1 - Tension-controlled sections as defined in 10.3.4 ....................................................0.90 (See also 9.3.2.7) 9.3.2.2 - Compression-controlled sections, as defined in 10.3.3: (a) Members with spiral reinforcement conforming to 10.9.3............................................0.75 (b) Other reinforced members .................................0.65 For sections in which the net tensile strain in the extreme tension steel at nominal strength, εt, is between the limits for compressioncontrolled and tension-controlled sections, ϕ shall be permitted to be linearly increased from that for compression-controlled sections to 0.90 as εt increases from the compression-controlled strain limit to 0.005. 9.3.2.3 - Shear and torsion .................................................0.75 9.3.2.4 - Bearing on concrete (except for post-tensioned anchorage zones and strut-and-tie models) ....................................0.65 9.3.2.5- Post-tensioned anchorage zones ………..…….....0.85

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3.4.1 Compression-Controlled Sections Sections are compression-controlled when the net tensile strain in the extreme tension steel εt is equal to or less than the compression-controlled strain limit at the time the concrete in compression reaches its assumed strain limit of 0.003. The compression-controlled strain limit is the net tensile strain in the reinforcement at balanced strain conditions. For Grade 420 reinforcement, and for all prestressed reinforcement, it is permitted to set the compression-controlled strain limit equal to 0.002.

3.4.2 Tension-Controlled Sections Sections are tension-controlled when the net tensile strain in the extreme tension steel is equal to or greater than 0.005 just as the concrete in compression reaches its assumed strain limit of 0.003. Sections with net tensile strain in the extreme tension steel between the compressioncontrolled strain limit and 0.005 constitute a transition region between compression-controlled and tension-controlled sections. 8

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Figure below shows the stress and strain conditions at the limit for tension-controlled sections. This limit is important because it is the limit for the use of ϕ = 0.9 (9.3.2.1). Critical parameters at this limit are given a subscript t.

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3.5 Design Assumptions 1. A plane section before bending remains plane after bending. 2. Stresses and strain are approximately proportional up to moderate loads (concrete stress 0.5 ́ ) . when the load is increased, the variation in the concrete stress is no longer linear. 3. Tensile strength of concrete is neglected in the design of reinforced concrete beams. 4. The maximum usable concrete compressive strain at the extreme fiber is assumed equal to 0.003 .

5. The steel is assumed to be uniformly strained to the strain that exists at the level of the centroid of the steel . Also if the strain in the steel is less than the yield strain of the steel , then the stress in the steel is . If , then the stress in steel will be equal to . 6. The bond between the steel and concrete is perfect and no slip occurs. Stress in concrete & reinforcement may be calculated from the strains using curves for concrete & steel. Compressive relationship for concrete may be assumed to be any shape (rectangular, trapezoidal, parabolic) that results in an acceptable prediction of strength.

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3.5.1 Equivalent Stress Distribution The actual distribution of concrete compressive stress is complex and usually not known explicitly. Research has shown that the important properties of concrete stress distribution can be approximated closely using any one of several different assumptions as to the form of stress distribution. The code permits any particular stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests. Many stress distributions have been proposed . The three most common are the parabola, trapezoid, and rectangle. The compressive force is modeled as Cc = k1k3f’c b*c at the location x = k2*c The compressive coefficients of the stress block at given for the following shapes. k3 is ratio of maximum stress at fc in the compressive zone of a beam to the cylinder strength, f c’ (0.85 is a typical value for common concrete).

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3.5.2 Whitney's Rectangular Stress distributions For purposes of simplification and practical application, a fictitious but equivalent rectangular concrete stress distribution was proposed. This rectangular stress distribution was proposed by Whitney (1942) and subsequently adopted by the ACI Code.

- According to figure above the average stress distribution is taken as : ́ - It is assumed to act over the upper area on the beam cross section defined by width b and depth a as shown in figures above. - The magnitude of a may determine by :

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where : c = distance from the outer fiber to the neutral axis = a factor dependent on concrete strength, and is given by ́ {

(

́

́

)

3.5.2.1 Requirements for Analysis of Reinforced Concrete Beams [1] Stress-Strain Compatibility : Stress at a point in member must correspond to strain at a point. [2] Equilibrium : Internal forces balances with external forces. 3.5.2.2 Rectangular Reinforced Concrete Beam ( Single Reinforced Section) (1) From forces Equilibrium. ⇒

∑

́ ∑

⇒

(

)

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(2) Find flexural capacity.

́

́ By taking a moment at compression force location

( (

)

́ (

)

)

́ (

)

́

(

́ (

For design use

́

(

√

)

́ ́

) )

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3.5.3 Flexural Failure Consequence There are three types of flexural failure of a structural member. - Steel may reach its yield strength before the concrete reaches its maximum. (Under-reinforced section).

Tension Failure

́

́

⇒

́

} - Concrete may fail before the yield of steel due to the presence of a high percentage of steel in the section. (Over-reinforced section). Compression Failure

⇒ }

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- Steel reaches yield at same time as concrete reaches ultimate strength. (Balanced section). :

Hence we simply equate the previous two equations: ́

́ }

3.5.4 Reinforcement Ratio Limitations

Which type of failure is the most desirable? The under-reinforced beam is the most desirable. 16

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fs = f y >> You want ductility →

system deflects and still carries load.

Although failure due to yielding of the steel is gradual with adequate warning of collapse, failure due to crushing of the concrete is sudden and without warning. The steel ratio (sometimes called reinforcement ratio) for rectangular section is given by:

[1] Maximum steel ratio: The under reinforced beam is preferred and ensured by the specifications of the ACI. - To ensure the section is always still as under reinforced section the ACI Code limited a minimum tensile strain from it we can derive a relation which give the maximum steel ratio which limited the section to be always still under reinforced section as : ́

Substitute

and ́

- The ACI Code further encourage the use of lower reinforcement ratios by allowing higher strength reduction factors in such sections. The Code defines a tension controlled member as having a net tensile strain greater than or equal to corresponding strength reduction factor is ϕ = 0.9 , so maximum reinforcement ratio for tension-controlled beam is: ́

Substitute

and 17

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́

[2] Minimum steel ratio The ACI Code establishes a lower limit on the amount of tension reinforcement. The code states that where tension reinforcement is required , the reinforcement area shall not less that given by : √́

Where - For a statically determinate beam with a flange in tension, the value of bw shall be the lesser of bf and 2bw. If As (provided) not required.

4/3 As (required) based on analysis, then As (min) is

4 3

 Mn  Mu - Temperature and Shrinkage reinforcement in structural slabs and footings ACI place perpendicular to direction of flexural reinforcement. For fy=280 or 300 MPa (GR 40 or GR 50 Bars): As (T&S) = 0.0020 Ag For fy=420 or 300 MPa (GR 60 or Welded Wire Fabric (WWF)): As (T&S) = 0.0018 Ag Ag - Gross area of the concrete.

- The steel ratio provided for sections shall be between max. and min. limits of ACI Code.

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3.5.5 SECTIONS WITH MULTIPLE LAYERS OF STEEL The simple and conservative way to design a beam with two layers of tension steel is to take d t equal to d, the depth to the centroid of all the tension steel. However, the code does permit the designer to take advantage of the fact that d t, measured to the center of the layer farthest from the compression face, is greater than d. The only time this would be necessary is when designing at or very close to the strain limit of 0.005 for tension-controlled sections. Therefore, the maximum steel ratio for beam with multiple layers of steel is: ́

́

3.5.6 Strength Reduction Factor ϕ Calculation of the nominal moment capacity frequently involves determination of the depth of equivalent rectangular stress block a . Since , it is some times more convenient to compute ratios than either or the net tensile strain. The assumption that p[lane sections remain plane ensure a direct correlation between net tensile and the ratios, as shown in Figure below( for Fy = 420 MPa ) .

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- For

Dr. AHMED AJEL

the section is tension control.

- For

and (Fy = 420 MPa) the section is compression control.

- In general

, with any value of Fy the section is

compression control. - For

between the range of tension control and compression control

(transition) the value of ϕ can be evaluated by linear interpolation for any value as: 1- For other sections (

) (

)

2- For spiral sections (

) (

)

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Example 3.2 Find the ACI design moment capacity use ́ and

of the beam shown in Figure, .

Solutions: (

)

Checking Steel Percentage

√́

√

́ ́

The section is Tension Control ( (

) )

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Example 3.3 Determine the ACI design moment capacity Figure, if ́ and

of the beam shown in .

Solutions: Checking Steel Percentage

√́

√

́ ́

́

⇒ The section is Tension Control

( (

) )

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Example 3.4 For the beam cross section shown in Figure, Calculate the ACI design ́ moment capacity , if (a) 20.7 MPa (b) 34.5 MPa (c) 62.1 MPa Solutions: (a) ́ Checking Steel Percentage

√́

√

́ ́

́

́

⇒ The section is Compression-Control . Hence the beam is not ductile and does not satisfy the ACI 318 Code. 23

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(b) ́ Checking Steel Percentage

√́

√

́ ́

́

́

Hence the beam is ductile, but in the transition zone with than 0.90. (

less

) (

(

) )

(

)

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( (

) )

(c) ́ Checking Steel Percentage

√́

√

́ ́

Hence the beam is ductile, tension controlled, with

= 0.90.

́

The section is tension control, ( (

.

) )

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3.5.7 Adequacy of Sections A given section is said to be adequate if the internal moment strength of the section is equal to or greater than the applied forced moment, or . The procedure can be summarized as follows: 1. Calculate the external applied factored moment,

2. calculate

for the basic singly reinforced section:

a. Check that b. Calculate c. Calculate 3. if

. ́

and check (

for .

).

, then the section is adequate .

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Assignment (2) H.W1 : An 2.44 m span cantilever beam has a rectangular section and reinforcement as shown in Figure. The beam carries a dead load , including its own weight of 2 kN.m and a live load of 1.22 kN.m. Using ́ , check if the beam is safe to carry the above loads.

H.W.2: A simply supported beam has a span of 6.1 m . If the cross ́ section of the beam is as shown in figure, , determine the allowable uniformly distributed service live load on the beam assuming the dead load that due to beam weight. Given : b = 305 mm , d= 432 mm, total depth h= 508 mm, and reinforced with three no. 8 bars (As = 1529 mm2).

H.W.3: Check the design adequacy of the section in Figure to resist a ́ factored moment Mu = 40.68 kN.m , using .

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H.W.4: Determine the design moment strength of a rectangular concrete section reinforced with four no. 9 bars in one row. Given : b = 305 mm , d= 419 mm, total depth h= 483 mm, and reinforced ́ with three no. 9 bars .

H.W.5: Determine the design moment strength of a rectangular concrete section reinforced with six no. 9 bars in two rows. Given : b = 305 mm , dt = 622 mm, total depth h= 686 mm, and ́ reinforced with three no. 9 bars .

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3.5.8 Details of Reinforcement Good reinforcement details are vital to satisfactory performance of reinforced concrete structures. Standard practice for reinforcing steel details has evolved gradually. The Building Code Committee (ACI 318) continually collects reports of research and practice related to structural concrete, suggests new research needed, and translates the results into specific code provisions for details of reinforcement. Recommended methods and standards for preparing design drawings, typical details, and drawings for the fabrication and placing of reinforcing steel in reinforced concrete structures are given in the ACI Detailing Manual, reported by ACI Committee 315.  Standard hooks The requirements for standard hooks for reinforcing bars are illustrated in Figures below shows the requirements for primary reinforcement and for stirrups and ties. The standard hook details for stirrups and ties apply to No. 8 and smaller bar sizes only. 7.1 — Standard hooks The term “standard hook” as used in this Code shall mean one of the following: 7.1.1 — 180-degree bend plus 4db extension, but not less than 65 mm at free end of bar. 7.1.2 — 90-degree bend plus 12db extension at free end of bar.

7.1.3 — For stirrup and tie hooks (a) No. 16 bar and smaller, 90-degree bend plus 6db extension at free end of bar; or (b) No. 19, No. 22, and No. 25 bar, 90-degree bend plus 12db extension at free end of bar; or (c) No. 25 bar and smaller, 135-degree bend plus 6db extension at free end of bar. 29

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 Minimum bend diameters Minimum bend diameter for a reinforcing bar is specifi ed as “the diameter of bend measured on the inside of the bar.” Minimum bend diameters, expressed as multiples of bar diameters, are dependent on bar size; for No.3 to No. 8 bars, the minimum bend diameter is 6 bar diameters; for No. 9 to No. 11 bars, the minimum bend diameter is 8 bar diameters; and for No. 14 and No. 18 bars, the minimum bend diameter is 10 bar diameters. Exceptions to these provisions are: 7.2 — Minimum bend diameters 7.2.1 — Diameter of bend measured on the inside of the bar, other than for stirrups and ties in sizes No. 10 through No. 16, shall not be less than the values in Table 7.2. 7.2.2 — Inside diameter of bend for stirrups and ties shall not be less than 4db for No. 16 bar and smaller. For bars larger than No. 16, diameter of bend shall be in accordance with Table 7.2. 7.2.3 — Inside diameter of bend in welded wire reinforcement for stirrups and ties shall not be less than 4db for deformed wire larger than MD40 and 2db for all other wires. Bends with inside diameter of less than 8db shall not be less than 4db from nearest welded intersection.

Bending

 Bending Reinforcement All reinforcement must be bent cold unless otherwise permitted by the licensed design professional. For unusual bends, special fabrication including heating may be required and the licensed design professional must give approval to the techniques used. 7.3 — Bending 7.3.1 — All reinforcement shall be bent cold, unless otherwise permitted by the licensed design professional. 7.3.2 — Reinforcement partially embedded in concrete shall not be field bent, except as shown on the design drawings or permitted by the licensed 31

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design professional.  Spacing Limits for Reinforcement The arrangement of bars within a beam must allow sufficient concrete on all sides of each bar to transfer forces into or out of bars , sufficient space so that the fresh concrete can be placed or consolidated around all the bars, and sufficient space to allow an internal vibrator to reach through to the bottom of the beam. Pencil type concrete immersion vibrators used in consolidation of the fresh concrete are (4-6 cm) in diameter. Enough space should be provided between the beam bars to allow a vibrator to reach the bottom of the form in at least one place in the beam width. ACI Code Sections 3.3.2, 7.6.1, and 7.6.2 specify the spacing and arrangements shown in figure below. 3.3.2 — Nominal maximum size of coarse aggregate shall be not larger than: (a) 1/5 the narrowest dimension between sides of forms, nor (b) 1/3 the depth of slabs, nor (c) 3/4 the minimum clear spacing between individual reinforcing bars or wires, bundles of bars, individual tendons, bundled tendons, or ducts. These limitations shall not apply if, in the judgment of the licensed design professional, workability and methods of consolidation are such that concrete can be placed without honeycombs or voids.

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7.6 — Spacing limits for reinforcement 7.6.1 — The minimum clear spacing between parallel bars in a layer shall be db, but not less than 25 mm. See also 3.3.2. 7.6.2 — Where parallel reinforcement is placed in two or more layers, bars in the upper layers shall be placed directly above bars in the bottom layer with clear distance between layers not less than 25 mm.

7.6.3 — In spirally reinforced or tied reinforced compression members, clear distance between longitudinal bars shall be not less than 1.5db nor less than 40 mm. See also 3.3.2. 7.6.4 — Clear distance limitation between bars shall apply also to the clear distance between a contact lap splice and adjacent splices or bars. 7.6.5 — In walls and slabs other than concrete joist construction, primary flexural reinforcement shall not be spaced farther apart than three times the wall or slab thickness, nor farther apart than 450 mm.

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 Bundled bars For isolated situations requiring heavy concentration of reinforcement, bundles of standard bar sizes can save space and reduce congestion for easier placement and consolidation of concrete. In those situations, bundling of bars in columns is a means to better locating and orienting the reinforcement for increased column capacity; also, fewer ties are required if column bars are bundled. Bundling of bars (parallel reinforcing bars in contact, assumed to act as a unit) is permitted, provided specific limitations are met. The limitations on the use of bundled bars are as follows: 7.6.6 — Bundled bars 7.6.6.1 — Groups of parallel reinforcing bars bundled in contact to act as a unit shall be limited to four in any one bundle. 7.6.6.2 — Bundled bars shall be enclosed within stirrups or ties. 7.6.6.3 — Bars larger than No. 36 shall not be bundled in beams.

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 Concrete Protection for Reinforcement It is necessary to have cover (concrete between the surface of the slab or beam and the reinforcement) for four primary reasons: 1. To bond the reinforcement to the concrete so that the two elements act together. The efficiency of the bond increases as the cover increase. A cover of at least one bar diameter is required for this purpose in beams and columns. 2. To protect the reinforcement against corrosion . 3. To protect the reinforcement from strength loss due to overheating in case of fire. The cover for fire protection is specified in the local building code. Generally speaking,20 mm cover to the reinforcement in structural slab will provide a 1 hour fire rating, while 40 mm cover to the stirrups or tie of beams corresponding to a 2-hour fire rating. 4. Additional cover sometimes is provided on the top of slabs, particularly in garages and factories, so that abrasion and wear due to traffic will not reduce the cover below that required for structural and other purposes. The amount of clear cover (is measured from the concrete surface to the outermost surface of the steel to which the cover requirement applies) will be based on ACI Code Section 7.7.1. 7.7 — Concrete protection for reinforcement 7.7.1 — Cast-in-place concrete (nonprestressed) Unless a greater concrete cover is required by 7.7.6 or 7.7.8, specified cover for reinforcement shall not be less than the following: Concrete cover, mm (a) Concrete cast against and permanently exposed to earth ............................................ 75 (b) Concrete exposed to earth or weather: No. 19 through No. 57 bars ..................................................50 No. 16 bar, MW200 or MD200 wire, and smaller…………...40 (c) Concrete not exposed to weather or in contact with ground: Slabs, walls, joists: No. 43 and No. 57 bars .........................................................40 34

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No. 36 bar and smaller.........................................................20 Beams, columns: Primary reinforcement, ties, stirrups, spirals ..................................................................................40 Shells, folded plate members: No. 19 bar and larger ..........................................................20 No. 16 bar, MW200 or MD200 wire, and smaller ………....13

Preliminary Beam and Slab Dimensions for Control of Deflections Estimations Beam Self Weight If the dimensions of the beam was chosen in analysis it can be used for calculation of the beam self- weight . The estimate the dimensions for self-weight can be by rules of thumb and engineering judgment. The weight of a rectangular beam will be about 15 % of the superimposed loads (dead, live, etc.). Assume b is about one-half of h . The preferred economical ratio between the beam depth and width b is between (2) and (3), if there is no architectural aspects . The deflections of a beam can be calculated from equations of the form

Rearranging this and making assumptions concerning strain distributions and neutral-axis depth eventually gives an equation of the form

Thus, for any acceptable ratio of deflection to span length, , it should be possible to specify span to-to-depth ratios, , which if exceeded may result in unacceptable deflections. It can be suggested that typical beam depths range between and , the selected beam depth , , will need to be checked against the minimum member thicknesses (depth, ) given in the second row of ACI Table 9.5 (a) for member not supporting partitions or other construction that are likely to be damaged by deflections.

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 Estimating the effective Depth of a Beam It is generally satisfactory to estimate the effective depth of a beam using the following approximations: For beams with one layer of reinforcement,

For beams with two layers of reinforcement,

For one-way slabs with spans up to 3.5m,

For one –way slabs with spans up to 3.5m,

It is important not to overestimate , because normal construction practice may lead to smaller value of than are shown on the drawings. Generally speaking , beam width b should not less than 25 cm , although with two bars, beams width as low as 18 cm can be used in extreme cases.

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3.5.9 Design The design of reinforced concrete members mean that the computations of proper sections dimensions and the amount of steel reinforcement required so that the structural member can resist the ultimate factored applied load safely. The ultimate strength for steel and concrete and the service loading should be, in some cases the all or some of section dimensions may be decided by architectural limits. The full design process accomplished by computing the required steel for stirrups to resist shear and torsion and checking the deflections and also checking development lengths and points of cut or bend for steel , all of this should be draw then on beam section .  Design of Beams when Section Dimensions are Known in this case ( )are known , and it only necessary to compute . This is actually a very common case for continuous members where the same size will be used in both positive and negative bending regions and may be used for several of the typical beam spans in a floor system. These dimensions may be established by architectural limits on member dimensions or may be established by designing the section of the beam that is resisting the largest bending moment. Example 3.5 Evaluate the steel area required for Simply Supported beam shown use bars with , the concrete strength ̇ . The beam span 5 m and carry loads (without self-weight), .

Solutions

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Dr. AHMED AJEL

(tension control section) ́

(

√

́

)

Assume one layer ⇒

(

Check if tension control (

√

)

)

⇒ The section is tension control, the assumption is correct. Check the steel limits √́

{ Use the maximum value O.K.

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⇒

No. of bars Check bar spacing (

(

)

)

O.K. Within the ACI Spacing limits

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Example 3.6 Find the necessary reinforcement for a given section, b = 380 mm, h = 510 mm. If it is subjected to a factored moment of 435 kN.m, the concrete strength ̇ and . Solutions: Assume

(tension control section) ́

(

√

́

)

Assume one layer ⇒

(

Check if tension control (

√

)

)

The section is in the transition zone.

Use # 32 bars No. of bars ⇒ ́

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Hence the beam is ductile, but in the transition zone with than 0.90. (

less

) (

(

) )

(

)

( (

) )

Check bar spacing: (

)

O.K.

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 Design of Beams when Section Dimensions are Not Known The second type of section design problem involves finding . Three decisions must be made here, that is, a preliminary estimate of the self-weight of the beam, selection of a target steel percentage, and final of ( ) the section dimensions Example 3.7 Design a rectangular beam for 10 m simple span to support a dead load of 20 kN/m (not including beam weight ) and live load of 30 kN/m . use , the concrete strength ̇ . Solutions: Assume that the beam weight is

Assume

.

(

) and use ( )

( )

́

( )

( ) ( )

From the equation below , find (

́ (

) ) 42

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224.32 We chose

then find

{

←

Use 500 mm x 800 mm section (d=680 mm) Beam weight

(

)(

)(

)

Use # 32 bars ⇒

No. of bars Check the steel limits √́

{ Use the maximum value

O.K.

Check bar spacing

(

)

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(

Dr. AHMED AJEL

)

O.K. Within the ACI Spacing limits Use six

32 bars in two rows (4914 mm2).

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